A) \[2\times {{10}^{-10}}\]
B) \[1.1\times {{10}^{-10}}\]
C) \[3.1\times {{10}^{-10}}\]
D) \[4.1\times {{10}^{-10}}\]
Correct Answer: B
Solution :
Key Idea: Find the relationship between \[{{K}_{sp}}\]and solubility after writing reaction for dissociation of \[AgCl\]. Now solve the problem. Given: solubility of\[AgCl=0.0015\,g/L=x\,g/L\] \[AgClA{{g}^{+}}+C{{l}^{-}}\] After dissociation \[x\] \[x\] \[x\] \[{{K}_{sp}}=A{{g}^{+}}]\,[C{{l}^{-}}]\] \[=x\times x\] \[{{K}_{sp}}={{x}^{2}}\] \[x=0.0015\,g/L=\frac{0.0015}{143.5}mol/L\]. \[\therefore \] \[{{K}_{sp}}{{\left( \frac{0.0015}{143.5} \right)}^{2}}\] \[=1.1\times {{10}^{-10}}\]
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